How do you find the volume of the solid bounded by the coordinate planes and the plane #5x + 5y + z = 6#?

Question

Luke Alvarez · Accepted Answer

#= 36/25#

this only makes sense if you are limiting it to the first octant.

in that case we can set #z = 0# and note that the plane and the x-y plane meet along the line #5x + 5y = 6# [which can also be written #x = 6/5 - y# or #y = 6/5 - x#]

The plane in the first octant is shaded yellow and intercepts shown

The volume we can get most simply from a double integral

#int_{y = 0}^{6/5} int_{x = 0}^{6/5 - y} dx dy qquad ( z(x,y) )#

#= int_{y = 0}^{6/5} int_{x = 0}^{6/5 - y} dx dy qquad (6 - 5x - 5y)#

#=int_{y = 0}^{6/5} dy qquad [ 6x - (5x^2)/2 - 5xy]_{x = 0}^{6/5 - y} #

#=int_{y = 0}^{6/5} dy qquad ( 6(6/5 - y) - (5(6/5 - y)^2)/2 - 5(6/5 - y)y ) #

#=1/10 int_{y = 0}^{6/5} dy qquad ( (6 - 5y)^2 ) #

# = 1/10 [- 1/15 (6 - 5y)^3 ]_{y = 0}^{6/5} #

# = 1/150 [ (6 - 5y)^3 ]_{y = 6/5}^{0} #

#= 36/25#

William Abdalla · Answer

undefined To find the volume of the solid bounded by the coordinate planes and the plane $5x + 5y + z = 6$, we can use the method of triple integration with appropriate limits.

1. Identify the region of integration.
2. Set up the triple integral.
3. Integrate with respect to each variable.

The given plane intersects the coordinate axes at points $(1, 0, 0)$, $(0, 1, 0)$, and $(0, 0, 6)$.

The region of integration is a triangular prism bounded by the coordinate planes and the plane $5x + 5y + z = 6$.

The volume $V$ is given by the triple integral:

$$V = \iiint_{D} \, dV$$

Where $dV$ represents an infinitesimal volume element, and $D$ is the region of integration.

The limits of integration for $x$, $y$, and $z$ are determined by the intersection points and the equation of the plane:

For $x$: $0 \leq x \leq 1$  
For $y$: $0 \leq y \leq 1$  
For $z$: $0 \leq z \leq 6 - 5x - 5y$

Now, integrate $dV = dx \, dy \, dz$ over the region $D$ using the given limits.

$$V = \int_0^1 \int_0^1 \int_0^{6-5x-5y} \, dz \, dy \, dx$$

$$V = \int_0^1 \int_0^1 (6 - 5x - 5y) \, dy \, dx$$

$$V = \int_0^1 \left[6y - 5xy - \frac{5}{2}y^2\right]_{0}^{1} \, dx$$

$$V = \int_0^1 (6 - 5x - \frac{5}{2}) \, dx$$

$$V = \left[6x - \frac{5}{2}x^2 - \frac{5}{2}x\right]_{0}^{1}$$

$$V = \left(6 - \frac{5}{2} - \frac{5}{2}\right) - \left(0 - 0 - 0\right)$$

$$V = 6 - 5 = 1$$

So, the volume of the solid bounded by the coordinate planes and the plane $5x + 5y + z = 6$ is $1$ cubic unit.